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Fronthaul Size: Calculation of maximum distance between RRH and BBU
April 01, 2014 | By Dr. Harrison J. Son (tech@netmanias.com), S.M. Shin (smshin@hfrnet.com)
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16

In C-RAN (Centralized/Cloud RAN), the more RRHs are covered by one CO (BBU hostelling site), i) the fewer COs are required, and ii) the more BBUs can be centralized at one CO. This allows for the most efficient utilization of BBU resources, thereby significantly reducing CAPEX/OPEX.

 

For this reason, the maximum separation distance between RRH and BBU should be secured. This distance is constrained by the timing requirement of Hybrid Automatic Retransmit reQuest (HARQ) protocol used as a retransmission mechanism between UE and eNB in an LTE network.

 

Figure 1. Latency requirement in C-RAN

 

As seen in Figure 1(a) above, according to this requirement, UE should receive ACK/NACK from eNB in three subframes after sending uplink data, i.e. in the fourth subframe. Otherwise, the UE retransmits the data.  

 

In LTE, uplink and downlink subsframes are typically time-aligned at eNB antenna port as shown in Figure 1(a). So, eNB should complete eNB processing (UL CPRI processing, UL frame decoding, ACK/NACK creation, DL frame creation, DL CPRI processing) within 3 msecs after receiving uplink data from UE in subframe n, and then send downlink ACK/NACK in subframe n+4 back to the UE.

 

Now, let's consider C-RAN where HARQ is processed between UE and BBU at the CO.

 

In usual C-RAN, BBU and RRH are located several kms, or even tens of kms, away from each other. So, additional delays like transmission delay via optical fiber, processing time at active equipment in a fronthaul network (e.g. Active WDM, PON, etc.) are caused while data are delivered from the antenna at a cell site to BBU at CO. The sum of these delays and baseband processing time at BBU must be less than 3 msecs.

 

In order to maintain the timing presented in Figure 1(a), the additional delay caused in the fronthaul network must be compensated somewhere, somehow, for example, by expediting the BBU processing as seen in Figure 1(b).

 

Base station vendors design BBU to complete the processing and send ACK/NACK usually within 2.75 msecs, intead of 3 msecs, in order to compensate the delay additionally caused in the fronthaul network in C-RAN. 

 

Therefore, about 250 μsecs can be allowed in the fronthaul network. Based on this delay budget, the maximum separation distance between BBU and RRH can be calculated (round-trip transmission latency of 10μs/Km).     

 

Table 1. Delay component at fronthaul (Uplink latency calculation)

 

In a fronthaul network built with Active WDM, delay components involved in the data transmission after RRH receives data from UE and before it sends ACK/NACK to UE are listed in Table 2 below.

 

In the table, delay components No. 1 through 3, which are caused at RRH and BBU, must be kept minimized by base station vendors while delay component no. 4 must be kept minimized by fronthaul vendors.

 

Table 2. Maximum fiber distance between RRH (cell site) and BBU (CO)

 

 

 

About Netmanias (www.netmanias.com)

NMC Consulting Group (Netmanias) is an advanced and professional network consulting company, specializing in IP network areas (e.g., FTTH, Metro Ethernet and IP/MPLS), service areas (e.g., IPTV, IMS and CDN), and wireless network areas (e.g., Mobile WiMAX, LTE and Wi-Fi) since 2002.

 

About HFR (www.hfrnet.com)

HFR has been actively responding to the Cloud RAN market under LTE environment. We expect that our
front-haul solution will become representative product in global equipment market. Also, HFR has been
leading the high-speed internet equipment with the development for Giga Internet service area. Based
on its competitive solutions in the wire and wireless communications fields, HFR is determined to become
Korea’s leading network equipment company.

 

 

Jesse Gomes 2014-07-03 23:22:55

Hi,

 

In the example

 

Maximum Fiber RTT = 3msec - (40usec + 10usec+2,7usec+4usec) = 243,3usec.

Netmanias 2014-07-15 11:07:59

Hi. Jesse

 

We assure the calculation result is correct as this:

 

Maximum Fiber RTT = 3msec - (40usec + 10usec+2.7msec+4usec)

                              = 3,000usec - (40usec + 10usec + 2,700usec + 4usec)

                              = 3,000usec - 2,754usec

                              = 246 usec

 

Thanks for your question.

Liu Jingchu 2014-08-27 12:03:11

Hi,

It seems you use 10us/Km = 1s/10e5 Km = 10e5 Km/s as the light speed. But the more accurate light speed in fibers would be 2 x 10e5 Km/s. Is there a conservative estimation or something?

Netmanias 2014-08-27 13:06:15

Hi. Liu

 

Let us answer your question as below:

 

10us/km is an abbreviated parameter to calculate an one-way "Maximum fiber distance" by multiplying a one-way transmission delay time per km(5usec/km) by 2. So, the above fomula could be re-expressed as "Maximum RTT / (2x5usec/km)".

 

Thus, the light speed over optic fiber can be calculated as "1/(one-way transmission delay time)= 1/(5usc/km) = 2x10E5km/s".

 

As you mentioned, 2x10E5km/s is the light speed over optic fiber cable.

 

I hope your question is asnwered by this comment.

 

Many questions from you will be welcomed. Thanks.

Dimitris 2014-10-05 08:14:33

Hi Netmanias,

 

Delay components (Table 2) do not take into account (d):Active equipment processing and (j):Fiber latency (BBU to RRH) of Table 1. Why & if this is right, what is the point then of having (d) & (j) shown in Table 1?

 

Tnx,

Dimitris

Netmanias 2014-10-07 12:11:15

Hi. Dimitris


Thank you for your question and comment on this article.


There was a mis-typing on the table as this: "c+k" should be "d+k".

We've corrected the mis-printed character directly now.

Sorry for making you confused by the error.


As you got a point, the fiber RTT can be calculated by subtracting a total of the delay components from 3msec. and then, you can estimate a maximum fiber distance (one-way) by dividing the calculated fiber RTT by 10us(=2x5us).


Thanks,

Netmanias


mehdi 2015-07-11 12:17:24

hello

according to :

Maximum Fiber distance (Km) = Maximum Fiber RTT / (10 µsec/Km)

in some article I studied that the distance is between 20 - 40 Km. can you please say how can I reach 40Km?

and what is the diffrent between PON 10 Gbps and PON 100 Gbps over this formula?

carlos 2015-07-13 23:30:10

Hi,

 

Eventually I'm missing something but I've one basic question:

 

In the example:

Maximum Fiber distance (Km) = Maximum Fiber RTT / (10 µsec/Km) = 24,6 Km

 

Since one is considering "round trip time" and taking in account Table2' variables, than "24,6 Km" should correspond to the trip from RRH (cell site) to BBU (CO) and back again to RRH (cell site), shouldn´t it? The "Maximum fiber distance between RRH (cell site) and BBU (CO)" shouldn´t be half of 24,6 Km, i.e. 12,3 Km?

 

Thanks.

 

BR,

Mehdi 2015-07-18 04:43:59

Hello carlos, 

according to "5 THINGS YOU SHOULD KNOW ABOUT FRONTHAUL" and "Cloud RAN for Mobile Networks - a Technology Overview" the distance is between 20-40Km

cheers.

Divya 2015-07-22 21:39:05

Hi,

 

3ms is the delay for how many bytes data (is it for a radio frame etc.)? Did you consider transmission delay which depends on the bandwidth of fronthaul? Your response is much appreciated. 

Dmitry Zuev 2016-12-05 23:19:40

Hi ,

What is about interoperability between BTS and transmission vendors with CPRI over Microwave case? 

Anyone has tried to integrate RRUS over E-band CPRI link with 6601 BTS? 

I see that some vendors like ALU allow to configure additional delay at BBU side to let CPRI synchronize over E-Band microwave link.

But I have got stuck with Ericsson as their BTS does not have an option.

Any advise ? 

Thanks,

Dmitry

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